$$\int_0^{2 \pi} \log\left(\frac{1}{2}[1+\sqrt{1-(a \sin\phi)^2}]\right) d\phi $$ have a closed form ?

An approximation for small $a$ is $2E-\pi$, but it is the exact form that is needed for any $|a|<1$.

The integration standing to Jack is

$$ -\frac{\pi a^2}{4}\phantom{}_4 F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;a^2\right) $$

The question now is: Can it be changed into a finite combination of elementary functions.

  • $\begingroup$ Are you sure about $2e-\pi$ for small $a$ ? This does not match what I obtain using numerical integration. Could you check if the formula is correct ?( $\endgroup$ – Claude Leibovici Mar 27 '15 at 10:35
  • $\begingroup$ Yes, it is E-pi. There is a missing parenthesis as mentionned by David. It is the integral from 0 to 2*pi of ln[0.5*(1+sqrt(1-square(a*sin(phi)))) ]. $\endgroup$ – MOEL Mar 27 '15 at 10:48
  • $\begingroup$ Sorry again, 2E-pi not E -pi. $\endgroup$ – MOEL Mar 27 '15 at 10:49
  • $\begingroup$ Is it possible to make the correction adding the missing parnethesis? $\endgroup$ – MOEL Mar 27 '15 at 10:51
  • $\begingroup$ Jack, E is not the euler number but the elliptic E(square(a)). $\endgroup$ – MOEL Mar 27 '15 at 10:54

For any $b\in(0,1)$ we have: $$I(b)=\int_{0}^{2\pi}\log\left(1+\sqrt{1-b^2\sin^2 x}\right)\,dx=4\int_{0}^{1}\frac{\log\left(1+\sqrt{1-b^2 t^2}\right)}{\sqrt{1-t^2}}\,dt$$ hence: $$ I(b)=\frac{1}{b}\int_{0}^{b}\frac{\log(1+\sqrt{1-u^2})}{\sqrt{1-\frac{u^2}{b^2}}}\,du=\frac{1}{b}\int_{0}^{\arcsin b}\frac{\cos\theta}{\sqrt{1-\frac{\sin^2\theta}{b^2}}}\,\log(1+\cos\theta)\,d\theta$$ and the last integral can be evaluated by exploiting the Fourier series of $\log(1+\cos\theta)$ that is pretty well-known. Another possible approach is differentiation under the integral sign.

We have: $$ I'(b) = -4\int_{0}^{\pi/2}\frac{b\sin^2 t}{\sqrt{1-b^2\sin^2 t}\left(1+\sqrt{1-b^2\sin^2 t}\right)}\,dt = \frac{2\pi-4K(b^2)}{b}\tag{1}$$ where $K$ is the complete elliptic integral of the first kind. Since $I(0)=2\pi\log 2$, it follows that:

$$ I(b)=2\pi\log 2-\frac{\pi b^2}{4}\phantom{}_4 F_3\left(1,1,\frac{3}{2},\frac{3}{2};2,2,2;b^2\right).$$

  • $\begingroup$ Jack, it is the excat form that is sought, not just changing the variables. However, the querry has been corrcted adding the parenthesis. $\endgroup$ – MOEL Mar 27 '15 at 11:19
  • $\begingroup$ @MOEL: that does not change anything. $\log(z/2)$ is just $\log z-\log 2$, hence get rid of the $-\log 2$ and take $b=a$. $\endgroup$ – Jack D'Aurizio Mar 27 '15 at 11:21
  • $\begingroup$ The variable change you have made is t=sin(x) ; this gives another form of the same integral but not the closed form. $\endgroup$ – MOEL Mar 27 '15 at 11:38
  • $\begingroup$ @MOEL: differentiation under the integral sign gives the closed form in terms of a hypergeometric function, too. See my updated answer. $\endgroup$ – Jack D'Aurizio Mar 27 '15 at 11:43
  • $\begingroup$ Why the downvote? What's wrong with this answer? $\endgroup$ – Jack D'Aurizio Mar 27 '15 at 16:06

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